FS2 7.1 - A normal mean with unknown variance

FS2 7.1 - A normal mean with unknown variance

An estimated standard deviation introduces uncertainty of its own. You will use the t distribution to test a normal population mean and construct confidence intervals when the population variance is unknown, using the unbiased sample variance and the correct tail.

Estimating the spread changes the reference distribution

If your estimated standard error is 1 mm1\,\mathrm{mm}, why can a 95%95\% confidence margin exceed 1.9600 mm1.9600\,\mathrm{mm}?

Because the standard error was estimated, rather than known. With 1010 observations, the appropriate t margin is 2.262 mm2.262\,\mathrm{mm}: about 15.4%15.4\% larger. The same-looking standard error carries an extra layer of uncertainty.

You already know

A known-variance normal interval uses xˉ±zα/2σ/n\bar x\pm z_{\alpha/2}\sigma/\sqrt n. The unbiased sample variance is s2=[∑x2−(∑x)2/n]/(n−1)s^2=[\sum x^2-(\sum x)^2/n]/(n-1); its square root ss estimates the population standard deviation. Confidence describes repeated-sampling coverage.

Here the independent observations come from one normal population with unknown variance. We estimate its standard error using s/ns/\sqrt n, then use a different sampling distribution.

Student's t statistic for a normal mean

For a random sample of n≥2n\ge2 observations from N(μ,σ2)\mathrm N(\mu,\sigma^2), with σ2>0\sigma^2>0, let S2S^2 be the unbiased sample variance. Then

T=Xˉ−μS/n∼tn−1T=\frac{\bar X-\mu}{S/\sqrt n}\sim t_{n-1}

The observed statistic uses ss, the positive square root of s2s^2. In the formula booklet.

The t distribution is symmetric about zero, with heavier tails than the standard normal distribution. As the degrees of freedom increase, its critical values approach the normal ones. No density proof is required here.

The booklet's t table gives upper-tail probabilities. For t9t_9, the upper 5%5\% point is 1.8331.833, and the upper 2.5%2.5\% point is 2.2622.262. Symmetry gives the corresponding negative lower points.

[DIAGRAM: asset_name: Estimated spread gives heavier reference tails; asset_slug: 9fm0-fs2-133-t-tails; caption: With nine degrees of freedom the central ninety-five per cent t cutoff exceeds the normal cutoff; description: Exact standard normal and Student t densities with nine degrees of freedom are drawn on common axes. The t curve has a lower central peak and heavier tails. Positive vertical guide lines mark normal cutoff 1.9600 and t cutoff 2.262 with labels placed on separate levels. A key identifies the curves and a note says that estimated rather than known spread requires the t procedure]
With nine degrees of freedom the central ninety-five per cent t cutoff exceeds the normal cutoff

The assumption concerns the original population, not a claim that every small sample will look perfectly bell-shaped. Under this model, use the t procedure for unknown variance rather than silently replacing it with a z calculation.

Key idea

Estimate spread with the unbiased s2s^2, use its square root in the standard error, and use n−1n-1 degrees of freedom for the t reference distribution.

Identify what is known before selecting a procedure.

Test the mean with a tail chosen from the claim

Under H0:μ=μ0H_0:\mu=\mu_0, calculate t=(xˉ−μ0)/(s/n)t=(\bar x-\mu_0)/(s/\sqrt n). A positive alternative uses the upper tail; a negative alternative uses the lower tail. For a two-sided alternative, allocate half the significance level to each tail.

These hypothesis and tail-selection rules are applications of the supplied sampling distribution, rather than separately printed booklet formulas. Choose the alternative before seeing the observed sign.

Worked example 1: a positive one-tailed test

Question. Ten independent component lengths come from a normal population with unknown variance. The measurements in millimetres give

∑x=520,∑x2=27130\sum x=520,\qquad \sum x^2=27130

Test at 5%5\% whether the population mean length exceeds 50 mm50\,\mathrm{mm}.

Plan. Find the unbiased spread, standardise the mean difference and compare with the upper 5%5\% point of t9t_9.

H0:μ=50,H1:μ>50xˉ=520/10=52s2=27130−5202/109=10unbiased variancesn=1010=1estimated standard errort=52−501=22>1.8339 degrees of freedom, upper 5%\begin{aligned} H_0&:\mu=50,\qquad H_1:\mu>50\\ \bar x&=520/10=52\\ s^2&=\frac{27130-520^2/10}{9}=10 &&\text{unbiased variance}\\ \frac{s}{\sqrt n}&=\sqrt{\frac{10}{10}}=1 &&\text{estimated standard error}\\ t&=\frac{52-50}{1}=2\\ 2&>1.833 &&9\text{ degrees of freedom, upper }5\% \end{aligned}

Why this step? The variance is unknown, so estimating its square root in the denominator requires a t cutoff, not a normal cutoff.

Answer. Reject H0H_0. There is sufficient evidence at 5%5\% that the population mean component length exceeds 50 mm50\,\mathrm{mm}.

Check. The statistic is positive, agreeing with the direction of the claimed increase.

Worked example 2 changes only the pre-specified question to whether the mean differs in either direction.

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